用于 memcpy、strcpy 和 strcpy_s 的 C6387

Question

看来还是无法摆脱C6387的警告

typedef struct HashBind{
    char* cKeyIdentifier;
    void* vValue;
} HashBind;

....
    
HashBind* strNewBind = malloc(sizeof(HashBind));    
strNewBind -> cKeyIdentifier = (char*) malloc((strlen(pcKey) + 1) * sizeof(char));
            
memcpy(strNewBind -> cKeyIdentifier, pcKey, strlen(pcKey + 1));

pcKey 是一个 const char* 类型。我怎样才能通过

Warning C6387 'strNewBind->cKeyIdentifier' could be '0': this does not adhere to the specification for the function 'memcpy'.

当我尝试使用 strcpy 或 strcpy_s 而不是 memcpy 时同样适用。任何想法或任何替代方案？如何跳过 strcpy/memcpy 的这种不安全使用（防止缓冲区溢出）？ C4496 and C6387 for using strcpy and strcat 没有太大帮助：/

Answer 1

'strNewBind->cKeyIdentifier' could be '0': this does not adhere to the specification for the function 'memcpy'.

测试来自 malloc() 的 NULL return。

size_t n = (strlen(pcKey) + 1) * sizeof(char);
strNewBind -> cKeyIdentifier = malloc(n);

// Add test
if (strNewBind->cKeyIdentifier) {            
  memcpy(strNewBind -> cKeyIdentifier, pcKey, n);
} else {
  Handle_OutOfMemory(); // TBD code.
}

用于 memcpy、strcpy 和 strcpy_s 的 C6387

C6387 for memcpy, strcpy and strcpy_s

c

warnings

memcpy

buffer-overflow

strcpy